∫(1-x^2)/[x(1+x^2)]dx
let
(1-x^2)/[x(1+x^2)] = a/x+ (b1x+b0)/(1+x^2)
1-x^2= a(1+x^2) +x(b1x+b0)
put x=0
a=1
coef.of x^2
-1=a+b1
b1=-2
coef.of x
b0=0
(1-x^2)/[x(1+x^2)] = 1/x- 2x/(1+x^2)
∫(1-x^2)/[x(1+x^2)]dx
=∫[1/x- 2x/(1+x^2)] dx
= ln|x| - ln(1+x^2) + C